Problem: Is ${367887}$ divisible by $3$ ?
Solution: A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {367887}= &&{3}\cdot100000+ \\&&{6}\cdot10000+ \\&&{7}\cdot1000+ \\&&{8}\cdot100+ \\&&{8}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {367887}= &&{3}(99999+1)+ \\&&{6}(9999+1)+ \\&&{7}(999+1)+ \\&&{8}(99+1)+ \\&&{8}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {367887}= &&\gray{3\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{8\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {3}+{6}+{7}+{8}+{8}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${367887}$ is divisible by $3$ if ${ 3}+{6}+{7}+{8}+{8}+{7}$ is divisible by $3$ Add the digits of ${367887}$ $ {3}+{6}+{7}+{8}+{8}+{7} = {39} $ If ${39}$ is divisible by $3$ , then ${367887}$ must also be divisible by $3$ Add the digits of ${39}$ $ {3}+{9} = \color{#9D38BD}{12} $ If $\color{#9D38BD}{12}$ is divisible by $3$ , then ${39}$ must also be divisible by $3$ $\color{#9D38BD}{12}$ is divisible by $3$, therefore ${367887}$ must also be divisible by $3$.